Problem: Factor the following expression: $-4$ $x^2+$ $3$ $x+$ $10$
Solution: This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-4)}{(10)} &=& -40 \\ {a} + {b} &=& & & {3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-40$ and add them together. Remember, since $-40$ is negative, one of the factors must be negative. The factors that add up to ${3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-5}$ and ${b}$ is ${8}$ $ \begin{eqnarray} {ab} &=& ({-5})({8}) &=& -40 \\ {a} + {b} &=& {-5} + {8} &=& 3 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {-4}x^2 {-5}x +{8}x +{10} $ Group the terms so that there is a common factor in each group: $ ({-4}x^2 {-5}x) + ({8}x +{10}) $ Factor out the common factors: $ x(-4x - 5) - 2(-4x - 5) $ Notice how $(-4x - 5)$ has become a common factor. Factor this out to find the answer. $(-4x - 5)(x - 2)$